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Definition: The elements of the set U6 = {z $\in$$\mathbb{C}$$|$ z6 = 1} are called the 6th roots of unity.

According to the definition of complex numbers, $\mathbb{C}$, we recall that z represents a + bi where a,b $\in \mathbb{R}$.

By letting U6 be the circle in the Euclidean plane with center at the origin and radius 1, we see that the elements of the set U6 are the numbers
cos $\left(m\frac{2\pi}{6}\right)$ + i sin $\left(m\frac{2\pi}{6}\right)$ for m = 0, 1, 2, 3, 4, 5.
Namely, the elements of this set are:

The image below represents these roots of unity on the unit circle.$$\zeta$0 is represented as z0, $$\zeta$1 is represented as z1, $$\zeta$2 is represented as z2, $$\zeta$3 is represented as z3, $$\zeta$4 is represented as z4, and $$\zeta$5 is represented as z5.

Geometrically, we multiply complex numbers, such as $$\zeta$0, $$\zeta$1, $$\zeta$2, $$\zeta$3, $$\zeta$4, $$\zeta$5, by multiplying their absolute values and adding their polar angles. The absolute values of the 6th roots of unity are all 1, and we know 1n = 1 for all n $\in \mathbb{Z}$. The polar angle between any 6th roots of unity is n$\left(\frac{\pi}{3}\right)$ for some n $\in \mathbb{Z}$ with addition modulo 2$\pi$, as can be seen in the above unit circle diagram. Since the multiplication of the absolute values of the 6th roots of unity are closed under multiplication, and the addition of the polar angles of the 6th roots of unity are closed under addition modulo 2$\pi$, then the elements in the set U6 are closed under complex multiplication.

Also, because $\zeta$6 = 1, we see these 6 powers of $\zeta$ are closed under multiplication. For example, we have

By the property of associativity in integer addition, we have $$\zeta$a+(b+c) = $$\zeta$(a+b)+c

= ($$\zeta$a+b) $\star$$$\zeta$c

($$\zeta$a$\star$$$\zeta$b) $\star$$$\zeta$c

Because for all $$\zeta$a, $$\zeta$b, $$\zeta$c$\in$U6 where a, b, c $\in \mathbb{Z}$ are arbitrary, we have $$\zeta$a$\star$ ($$\zeta$b$\star$$$\zeta$c) = ($$\zeta$a$\star$$$\zeta$b) $\star$$$\zeta$c, associativity holds, meeting the first group axiom.

Also from the Introduction to the Group section, we know that an element in the set U6 is (cos 0 + i sin 0) = 1 = $$\zeta$0. As in the sets $\mathbb{R}$ and $\mathbb{Z}$ with multiplication, the identity element for $\mathbb{C}$ with multiplication is the number 1. Since the elements in U6 are complex numbers and the operation is complex multiplication, the identity element for this group is 1 also, or to be more notationally correct the identity element in U6 is $$\zeta$0. For all a $\in \mathbb{Z}$ is arbitrary and therefore all $$\zeta$a$\in$U6 be arbitrary, we have

$$\zeta$0$\star$$$\zeta$a = $$\zeta$0+a = $$\zeta$a

AND $$\zeta$a$\star$$$\zeta$0 = $$\zeta$a+0 = $$\zeta$a, showing that the identity element is $$\zeta$0 for $\star$.

We show that each element in U6, namely ($$\zeta$0, $$\zeta$1, $$\zeta$2, $$\zeta$3, $$\zeta$4, and $$\zeta$5) has an inverse in U6 equaling the identity element $\zeta$0.

For element $\zeta$0 in U6, there is the inverse element $\zeta$0 in U6 such that $\zeta$0$\star$$$\zeta$0 = $\zeta$0+0 = $\zeta$0, the identity element.

For element $\zeta$1 in U6, there is the inverse element $\zeta$5 in U6 such that $\zeta$1$\star$$\zeta$5 = $\zeta$1+5 = $\zeta$6 = 1 = $\zeta$0, the identity element.

For element $\zeta$2 in U6, there is the inverse element $\zeta$4 in U6 such that $\zeta$2$\star$$\zeta$4 = $\zeta$2+4 = $\zeta$6 = 1 = $\zeta$0, the identity element.

For element $\zeta$3 in U6, there is the inverse element $\zeta$3 in U6 such that $\zeta$3$\star$$\zeta$3 = $\zeta$3+3 = $\zeta$6 = 1 = $\zeta$0, the identity element.

For element $\zeta$4 in U6, there is the inverse element $\zeta$2 in U6 such that $\zeta$4$\star$$\zeta$2 = $\zeta$4+2 = $\zeta$6 = 1 = $\zeta$0, the identity element.

For element $\zeta$5 in U6, there is the inverse element $\zeta$1 in U6 such that $\zeta$5$\star$$\zeta$1 = $\zeta$5+1 = $\zeta$6 = 1 = $\zeta$0, the identity element.

Because all three group axioms are satisfied, we have shown that (U6, $\star$) is indeed a group.

It is obvious by the definition of a subgroup that (U6, $\star$) $\leq$ (U6, $\star$) as U6$\subseteq$U6 and (U6, $\star$) is a group itself.

Similarly, U3$\subset$U6, as can be seen by comparing the graph below with the graph in the Introduction to the Group section, and (U3, $\star$) is a group itself.
Therefore by the definition of a subgroup, (U3, $\star$) $\leq$ (U6, $\star$).

And lastly, U1$\subset$U6, and (U1, $\star$) is a group itself.
Therefore by the definition of a subgroup, (U1, $\star$) $\leq$ (U6, $\star$).